Statistics Class 10 Mathematics Notes

Introduction to Statistics

Statistics is the science of collecting, organizing, analyzing, and interpreting data. Data can be in the form of numbers, measurements, observations, or descriptions. The goal of statistics is to make sense of this data and draw meaningful conclusions.

In Class IX, you studied ungrouped data and grouped data.

Ungrouped data is a raw collection of individual data values, like a list of students' marks.
Grouped data is data that has been organized into classes or intervals, along with their corresponding frequencies. For example, age groups (0-10, 10-20) with the number of people in each group.

You also learned to represent data using:

Bar graphs: Used for discrete data, showing categories with rectangular bars.
Histograms: Used for continuous data, where bars touch each other, representing frequency within intervals.
Frequency polygons: Line graphs connecting the midpoints of histogram bars to show the shape of the distribution.

Furthermore, you explored measures of central tendency for ungrouped data. These are single values that represent the "center" or typical value of a dataset. The main measures are:

Mean: The average of all data points.
Median: The middle value when data is arranged in order.
Mode: The value that appears most frequently.

In this chapter, we will extend these concepts to grouped data. This is essential because real-world data is often very large and complex, making it impractical to work with as individual values. Grouping data allows us to summarize and analyze it more efficiently.

We will also learn about:

Cumulative frequency: The sum of frequencies for all data values up to a certain point.
Cumulative frequency distribution: A table showing cumulative frequencies.
Ogives: Graphs of cumulative frequency distributions, used to find median, quartiles, etc.

Measures of Central Tendency for Grouped Data

When data is grouped, we cannot calculate the exact mean, median, or mode as we did for ungrouped data. Instead, we use formulas that give us an estimated or approximate value for these measures. These approximations are very useful for understanding the central tendency of large datasets.

The three main measures of central tendency for grouped data that we will study are:

Measure	What it Represents	How it's Calculated for Grouped Data
Mean	The average value of the data	Estimated using formulas
Median	The middle value of the data	Estimated using formulas
Mode	The most frequent value in the data	Estimated using formulas

The methods for calculating these for grouped data are more advanced and will be covered in detail in subsequent sections. This introduction sets the stage by reminding us of the foundational concepts from Class IX and highlighting the need to adapt these measures for grouped data.

Remember
Statistics helps us make sense of data by summarizing it using measures of central tendency and visual representations. Grouped data requires specific formulas to estimate these measures. 🎯 Exam-smart: Questions might ask you to identify what statistics is about or to recall the types of graphs used for data representation. Understanding the difference between grouped and ungrouped data is also a common starting point.

Mean of Grouped Data

When data is presented in the form of a frequency distribution table with class intervals, we often need to find the mean to understand the central tendency of the data. Calculating the exact mean for grouped data can be complex, so we use approximations based on the assumption that data within each class interval is uniformly distributed or concentrated at the class mark.

There are three primary methods to calculate the mean of grouped data:

1. Direct Method

This method assumes that all values within a class interval are equal to its class mark (mid-point).

Class Mark (xᵢ): The midpoint of a class interval, calculated as $\frac{Upper limit + Lower limit}{2}$ .
Formula:

\overset{ˉ}{x} = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}

where $f_{i}$ is the frequency of the $i$ -th class and $x_{i}$ is the class mark of the $i$ -th class.

Example 1 (Ungrouped to Grouped Data)

Given: Marks obtained by 30 students in a Mathematics paper.

Marks Obtained	Number of Students ( $f_{i}$ )
10	1
20	1
36	3
40	4
50	3
56	2
60	4
70	4
72	1
80	1
88	2
92	3
95	1

To find: The mean of the marks obtained.

Solution:

First, we calculate the product $f_{i} x_{i}$ for each class.

Marks Obtained ( $x_{i}$ )	Number of Students ( $f_{i}$ )	$f_{i} x_{i}$
10	1	10
20	1	20
36	3	108
40	4	160
50	3	150
56	2	112
60	4	240
70	4	280
72	1	72
80	1	80
88	2	176
92	3	276
95	1	95
Total	$\sum f_{i} = 30$	$\sum f_{i} x_{i} = 1779$

\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{1779}{30} = 59.3

Answer: The mean marks obtained by the students is 59.3.

Example 2 (Grouped Data)

Given: A grouped frequency distribution of marks.

Class Interval	Number of Students ( $f_{i}$ )
10 - 25	2
25 - 40	3
40 - 55	7
55 - 70	6
70 - 85	6
85 - 100	6

To find: The mean of the marks using the Direct Method.

Solution:

First, we find the class mark ( $x_{i}$ ) for each interval.

Class Interval	Number of Students ( $f_{i}$ )	Class Mark ( $x_{i}$ )	$f_{i} x_{i}$
10 - 25	2	17.5	35.0
25 - 40	3	32.5	97.5
40 - 55	7	47.5	332.5
55 - 70	6	62.5	375.0
70 - 85	6	77.5	465.0
85 - 100	6	92.5	555.0
Total	$\sum f_{i} = 30$		$\sum f_{i} x_{i} = 1860.0$

\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{1860.0}{30} = 62

Answer: The mean of the marks is 62.

Remember
The Direct Method gives the exact mean if the data is already grouped with class marks. For data where frequencies are assumed to be at class marks, it provides an approximate mean. The result of Example 1 (59.3) and Example 2 (62) differ because in Example 2, we used class marks as representatives, making it an approximation.

Exam-smart
In case/source-based questions, you might be given a frequency table and asked to calculate the mean. You'll need to identify the class marks, calculate $f_{i} x_{i}$ , sum them up, and divide by the sum of frequencies. Be mindful if the question specifies "exact" or "approximate" mean.

2. Assumed Mean Method

This method simplifies calculations when $x_{i}$ and $f_{i}$ are large by choosing an assumed mean (a), usually the class mark of a central class.

Assumed Mean (a): A value chosen from the $x_{i}$ values, typically the middle one.
Deviation ( $d_{i}$ ): The difference between each class mark and the assumed mean: $d_{i} = x_{i} - a$ .
Formula:

\overset{ˉ}{x} = a + \frac{\sum_{i = 1}^{n} f_{i} d_{i}}{\sum_{i = 1}^{n} f_{i}}

Example 2 (Assumed Mean Method)

Given: The same grouped data as in Example 2.

Class Interval	Number of Students ( $f_{i}$ )	Class Mark ( $x_{i}$ )
10 - 25	2	17.5
25 - 40	3	32.5
40 - 55	7	47.5
55 - 70	6	62.5
70 - 85	6	77.5
85 - 100	6	92.5

To find: The mean using the Assumed Mean Method.

Solution:

Let's choose the class mark of the central class (40-55) as the assumed mean, so $a = 47.5$ .

Class Interval	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 47.5$	$f_{i} d_{i}$
10 - 25	2	17.5	-30	-60
25 - 40	3	32.5	-15	-45
40 - 55	7	47.5	0	0
55 - 70	6	62.5	15	90
70 - 85	6	77.5	30	180
85 - 100	6	92.5	45	270
Total	$\sum f_{i} = 30$			$\sum f_{i} d_{i} = 435$

\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}} = 47.5 + \frac{435}{30} = 47.5 + 14.5 = 62

Answer: The mean of the marks is 62.

Remember
The Assumed Mean Method is useful when the values of $x_{i}$ are large. Choosing a central value for 'a' minimizes the magnitudes of $d_{i}$ , simplifying calculations. The mean calculated will be the same regardless of the chosen 'a'.

Exam-smart
If given large numbers in $x_{i}$ or $f_{i}$ , consider the Assumed Mean Method. You'll need to identify a suitable 'a', calculate deviations $d_{i}$ , compute $f_{i} d_{i}$ , and then apply the formula. Watch out for calculation errors with negative numbers.

3. Step-Deviation Method

This method further simplifies calculations by introducing a new variable ' $u_{i}$ ', which is the deviation divided by the class size. It's most effective when all deviations ( $d_{i}$ ) have a common factor, ideally the class size (h).

Class Size (h): The difference between the upper and lower limits of a class interval (assuming it's constant for all intervals).
Step Deviation ( $u_{i}$ ): $u_{i} = \frac{d_{i}}{h} = \frac{x_{i} - a}{h}$ .
Formula:

\overset{ˉ}{x} = a + (\frac{\sum_{i = 1}^{n} f_{i} u_{i}}{\sum_{i = 1}^{n} f_{i}}) \times h

Example 2 (Step-Deviation Method)

Given: The same grouped data as in Example 2.

Class Interval	Number of Students ( $f_{i}$ )	Class Mark ( $x_{i}$ )
10 - 25	2	17.5
25 - 40	3	32.5
40 - 55	7	47.5
55 - 70	6	62.5
70 - 85	6	77.5
85 - 100	6	92.5

To find: The mean using the Step-Deviation Method.

Solution:

From the Assumed Mean Method, we know $a = 47.5$ . The class size $h = 25 - 10 = 15$ .

Class Interval	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 47.5$	$u_{i} = \frac{d_{i}}{15}$	$f_{i} u_{i}$
10 - 25	2	17.5	-30	-2	-4
25 - 40	3	32.5	-15	-1	-3
40 - 55	7	47.5	0	0	0
55 - 70	6	62.5	15	1	6
70 - 85	6	77.5	30	2	12
85 - 100	6	92.5	45	3	18
Total	$\sum f_{i} = 30$				$\sum f_{i} u_{i} = 29$

\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h = 47.5 + (\frac{29}{30}) \times 15 = 47.5 + 29 \times \frac{1}{2} = 47.5 + 14.5 = 62

Answer: The mean of the marks is 62.

Remember
The Step-Deviation Method is the most efficient when class sizes are equal and the deviations ( $d_{i}$ ) share a common factor. It significantly reduces the size of numbers involved in calculations.

Exam-smart
For problems with equal class sizes where $d_{i}$ values are large multiples of the class size, the Step-Deviation Method is ideal. You'll need to correctly identify 'a', 'h', calculate $u_{i}$ values, then $f_{i} u_{i}$ , and apply the formula carefully.

Comparison of Methods for Mean

Feature	Direct Method	Assumed Mean Method	Step-Deviation Method
When to Use	Small $x_{i}$ and $f_{i}$ values.	Large $x_{i}$ or $f_{i}$ values.	Large $x_{i}$ and $f_{i}$ with equal class sizes that share common factor.
Simplicity	Simplest conceptually, more calculation.	Reduces calculation size.	Minimizes calculation size.
Formula	$\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$	$\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$	$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$
Variables	$f_{i}, x_{i}$	$f_{i}, x_{i}, a, d_{i} = x_{i} - a$	$f_{i}, x_{i}, a, d_{i}, u_{i} = \frac{d_{i}}{h}, h$

Example 2 (All Three Methods)

Given: Percentage distribution of female teachers.

Percentage of female teachers	Number of States/U.T. ( $f_{i}$ )
15 - 25	6
25 - 35	11
35 - 45	7
45 - 55	4
55 - 65	4
65 - 75	2
75 - 85	1

To find: Mean percentage by all three methods.

Solution:

First, calculate class marks ( $x_{i}$ ). $h = 10$ . Let's choose $a = 50$ .

Class Interval	$f_{i}$	$x_{i}$	$f_{i} x_{i}$	$d_{i} = x_{i} - 50$	$u_{i} = d_{i} / 10$	$f_{i} d_{i}$	$f_{i} u_{i}$
15 - 25	6	20	120	-30	-3	-180	-18
25 - 35	11	30	330	-20	-2	-220	-22
35 - 45	7	40	280	-10	-1	-70	-7
45 - 55	4	50	200	0	0	0	0
55 - 65	4	60	240	10	1	40	4
65 - 75	2	70	140	20	2	40	4
75 - 85	1	80	80	30	3	30	3
Total	$\sum f_{i} = 35$		$\sum f_{i} x_{i} = 1400$			$\sum f_{i} d_{i} = - 360$	$\sum f_{i} u_{i} = - 36$

Direct Method:

\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{1400}{35} = 40

(Note: The text provided stated $\sum f_{i} x_{i} = 1390$ . Recalculating: $6 (20) + 11 (30) + 7 (40) + 4 (50) + 4 (60) + 2 (70) + 1 (80) = 120 + 330 + 280 + 200 + 240 + 140 + 80 = 1390$ . So $\overset{ˉ}{x} = 1390 / 35 = 39.71$ )

Assumed Mean Method:

\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}} = 50 + \frac{- 360}{35} = 50 - 10.2857 \approx 39.71

Step-Deviation Method:

\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h = 50 + (\frac{- 36}{35}) \times 10 = 50 - 10.2857 \approx 39.71

Answer: The mean percentage of female teachers is approximately 39.71%.

Remember
All three methods yield the same result if calculations are done correctly, especially when dealing with approximations. The choice of method depends on the nature of the data.

Exam-smart
Questions may ask you to use a specific method or the most appropriate method. Understand the criteria for choosing each method to save time and reduce errors.

Example 3 (Unequal Class Sizes)

Given: Distribution of wickets taken by bowlers.

Number of Wickets	Number of Bowlers ( $f_{i}$ )
20 - 60	7
60 - 100	5
100 - 150	16
150 - 250	12
250 - 350	2
350 - 450	3

To find: Mean number of wickets.

Solution:

Here, class sizes are unequal. The Step-Deviation Method can still be used if we select 'h' as a divisor of all $d_{i}$ . Let's choose $a = 200$ and $h = 20$ (a common factor for many $d_{i}$ ).

Number of Wickets	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 200$	$u_{i} = d_{i} / 20$	$f_{i} u_{i}$
20 - 60	7	40	-160	-8	-56
60 - 100	5	80	-120	-6	-30
100 - 150	16	125	-75	-3.75	-60
150 - 250	12	200	0	0	0
250 - 350	2	300	100	5	10
350 - 450	3	400	200	10	30
Total	$\sum f_{i} = 45$				$\sum f_{i} u_{i} = - 106$

\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h = 200 + (\frac{- 106}{45}) \times 20 = 200 - \frac{2120}{45} \approx 200 - 47.11 = 152.89

Answer: The mean number of wickets taken is approximately 152.89. This signifies that, on average, bowlers in this distribution took about 152.89 wickets.

Remember
When class sizes are unequal, the Step-Deviation Method can still be applied by choosing a suitable value for 'h' that divides most of the deviations. The Direct Method is always applicable but can be computationally intensive.

Exam-smart
Questions with unequal class intervals are designed to test your flexibility. You might need to adapt the Step-Deviation Method by picking an appropriate 'h', or revert to the Assumed Mean Method if $d_{i}$ values are hard to divide uniformly.

13.3 Mode of Grouped Data

The mode of a data set is the value that occurs most frequently. For grouped data, we first identify the modal class, which is the class with the highest frequency. The mode is then estimated using a formula.

Modal Class: The class interval with the maximum frequency.
Formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h

where:

$l$ = lower limit of the modal class
$h$ = size of the class interval (assuming all are equal)
$f_{1}$ = frequency of the modal class
$f_{0}$ = frequency of the class preceding the modal class
$f_{2}$ = frequency of the class succeeding the modal class

Example 4 (Ungrouped Data)

Given: Wickets taken by a bowler in 10 matches: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3.

To find: The mode.

Solution: Let's create a frequency table:

Number of Wickets	Frequency
0	1
1	1
2	3
3	2
4	1
5	1
6	1

The maximum frequency is 3, which corresponds to the number of wickets being 2.

Answer: The mode is 2.

Example 5 (Grouped Data)

Given: Frequency distribution of family members in households.

Family Size	Number of Families ( $f$ )
1 - 3	7
3 - 5	8
5 - 7	2
7 - 9	2
9 - 11	1

To find: The mode.

Solution: The maximum frequency is 8, so the modal class is 3 - 5.

$l = 3$ (lower limit of modal class)
$h = 2$ (class size)
$f_{1} = 8$ (frequency of modal class)
$f_{0} = 7$ (frequency of preceding class)
$f_{2} = 2$ (frequency of succeeding class)

Mode = 3 + (\frac{8 - 7}{2 \times 8 - 7 - 2}) \times 2 = 3 + (\frac{1}{16 - 7 - 2}) \times 2 = 3 + (\frac{1}{7}) \times 2 = 3 + \frac{2}{7} \approx 3 + 0.286 = 3.286

Answer: The mode of the data is approximately 3.286. This suggests that the most common family size is around 3.286 members.

Example 6 (Mode and Mean Comparison)

Given: Marks distribution of 30 students (from Example 1's grouped data).

Class Interval	Number of Students ( $f$ )
10 - 25	2
25 - 40	3
40 - 55	7
55 - 70	6
70 - 85	6
85 - 100	6

To find: Mode and compare it with the mean (from Example 1, which was 62).

Solution: The maximum frequency is 7, so the modal class is 40 - 55.

$l = 40$
$h = 15$
$f_{1} = 7$
$f_{0} = 3$
$f_{2} = 6$

Mode = 40 + (\frac{7 - 3}{2 \times 7 - 3 - 6}) \times 15 = 40 + (\frac{4}{14 - 9}) \times 15 = 40 + (\frac{4}{5}) \times 15 = 40 + 4 \times 3 = 40 + 12 = 52

Answer: The mode is 52. Interpretation: The mean marks obtained by students is 62, while the most common marks obtained by a significant number of students lie in the modal class, with the calculated mode being 52. This indicates that while the average performance is 62, the most frequent performance cluster is around 52.

Remember
The mode is directly influenced by the highest frequency, whereas the mean is influenced by all data values. They represent different aspects of the data's central tendency.

Exam-smart
When asked to compare mean and mode, be prepared to calculate both and then state what each signifies. For instance, "The mean is 62, representing the average score, while the mode is 52, representing the most frequently scored marks." This comparison helps understand the distribution's skewness.

Median of Grouped Data

The median is a measure of central tendency that represents the middle value in a dataset. For grouped data, which is presented in intervals, we cannot simply pick the middle observation as the data is not listed individually. Instead, we find the class interval that contains the middle observation and then use a formula to estimate the exact median value within that interval. This helps us understand the central point of the data distribution.

To find the median of grouped data, we first need to construct a cumulative frequency table. A cumulative frequency for a class interval represents the total frequency of all observations that fall within that class interval or any preceding class interval.

There are two types of cumulative frequency distributions:

Less than type: This counts the number of observations with values less than the upper limit of each class.
More than type: This counts the number of observations with values greater than or equal to the lower limit of each class.

Finding the Median Class

Once we have the cumulative frequency table (typically the 'less than' type is used for median calculation), we determine the total number of observations, denoted by $n$ . The median value will lie in the class interval where the cumulative frequency crosses the value $\frac{n}{2}$ .

We calculate $\frac{n}{2}$ .
We locate the median class, which is the class interval whose cumulative frequency is the first one greater than or equal to $\frac{n}{2}$ .

Formula for Median of Grouped Data

The median of grouped data is calculated using the formula:

Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h

Where:

$l$ is the lower limit of the median class.
$n$ is the total number of observations (the sum of all frequencies).
$cf$ is the cumulative frequency of the class preceding the median class.
$f$ is the frequency of the median class.
$h$ is the class size (the difference between the upper and lower limits of a class interval).

This formula estimates the median by assuming that the values within the median class are uniformly distributed. $l$ gives us a starting point, and the term $(\frac{\frac{n}{2} - cf}{f}) \times h$ adjusts this starting point based on how many observations we need to pass within the median class to reach the $\frac{n}{2}$ mark.

Example 1: Calculating Median from a Grouped Frequency Distribution

Given: The following data shows the marks obtained by 100 students in a test:

Marks obtained	Number of students (f)
0 - 10	5
10 - 20	3
20 - 30	4
30 - 40	3
40 - 50	3
50 - 60	4
60 - 70	7
70 - 80	9
80 - 90	7
90 - 100	8

To find: The median marks.

Solution: First, we create the cumulative frequency table:

Marks	Frequency (f)	Cumulative Frequency (cf)
0 - 10	5	5
10 - 20	3	8
20 - 30	4	12
30 - 40	3	15
40 - 50	3	18
50 - 60	4	22
60 - 70	7	29
70 - 80	9	38
80 - 90	7	45
90 - 100	8	53

Total number of observations, $n = 53$ .
Calculate $\frac{n}{2} = \frac{53}{2} = 26.5$ .
The median class is the class where the cumulative frequency first exceeds 26.5. Looking at the 'Cumulative Frequency' column, the class 60 - 70 has a cumulative frequency of 29, which is the first value greater than 26.5.
So, the median class is 60 - 70.

Now, we identify the values for the median formula:

$l$ (lower limit of the median class) = 60
$n$ = 53
$cf$ (cumulative frequency of the class preceding 60 - 70, which is 50 - 60) = 22
$f$ (frequency of the median class 60 - 70) = 7
$h$ (class size) = 10

Substitute these values into the formula:

Median = 60 + (\frac{26.5 - 22}{7}) \times 10

Median = 60 + (\frac{4.5}{7}) \times 10

Median = 60 + \frac{45}{7}

Median = 60 + 6.428...

Median \approx 66.43

Answer: The median marks are approximately 66.43. This means that about half the students scored less than 66.43 marks, and the other half scored more.

Example 2: Finding Unknown Frequencies using Median

Given: The median of the following distribution is 525. The total frequency is 100.

Class interval	Frequency
0 - 100	2
100 - 200	5
200 - 300	x
300 - 400	12
400 - 500	17
500 - 600	20
600 - 700	y
700 - 800	9
800 - 900	7
900 - 1000	4

To find: The values of $x$ and $y$ .

Solution: First, we create the cumulative frequency table:

Class interval	Frequency (f)	Cumulative Frequency (cf)
0 - 100	2	2
100 - 200	5	7
200 - 300	x	$7 + x$
300 - 400	12	$19 + x$
400 - 500	17	$36 + x$
500 - 600	20	$56 + x$
600 - 700	y	$56 + x + y$
700 - 800	9	$65 + x + y$
800 - 900	7	$72 + x + y$
900 - 1000	4	$76 + x + y$

The total frequency is given as $n = 100$ .
From the cumulative frequency column, we have $76 + x + y = 100$ . This simplifies to the equation: $x + y = 24$ (Equation 1).
The median is given as 525. Since 525 lies between 500 and 600, the median class is 500 - 600.

Now, we identify the values for the median formula:

$l$ (lower limit of the median class) = 500
$n$ = 100
$\frac{n}{2} = \frac{100}{2} = 50$ .
$cf$ (cumulative frequency of the class preceding 500 - 600, which is 400 - 500) = $36 + x$ .
$f$ (frequency of the median class 500 - 600) = 20.
$h$ (class size) = 100.

Substitute these values into the median formula:

525 = 500 + (\frac{50 - (36 + x)}{20}) \times 100

525 - 500 = (\frac{50 - 36 - x}{20}) \times 100

25 = (\frac{14 - x}{20}) \times 100

Divide both sides by 100:

\frac{25}{100} = \frac{14 - x}{20}

0.25 = \frac{14 - x}{20}

Multiply both sides by 20:

0.25 \times 20 = 14 - x

5 = 14 - x

x = 14 - 5

x = 9

Now substitute the value of $x$ into Equation 1 ( $x + y = 24$ ):

9 + y = 24

y = 24 - 9

y = 15

Answer: $x = 9$ and $y = 15$ .

Example 3: Median from "Less Than" Type Data

Given: The following data represents the heights of 51 girls:

Height (in cm)	Number of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

To find: The median height.

Solution: First, we convert the "less than" type data into class intervals and find their frequencies:

Class intervals	Frequency (f)	Cumulative Frequency (cf)
Below 140	4	4
140 - 145	$11 - 4 = 7$	11
145 - 150	$29 - 11 = 18$	29
150 - 155	$40 - 29 = 11$	40
155 - 160	$46 - 40 = 6$	46
160 - 165	$51 - 46 = 5$	51

Total number of observations, $n = 51$ .
Calculate $\frac{n}{2} = \frac{51}{2} = 25.5$ .
The median class is the class where the cumulative frequency first exceeds 25.5. Looking at the 'Cumulative Frequency' column, the class 145 - 150 has a cumulative frequency of 29, which is the first value greater than 25.5.
So, the median class is 145 - 150.

Identify the values for the median formula:

$l$ (lower limit of the median class) = 145
$n$ = 51
$cf$ (cumulative frequency of the class preceding 145 - 150, which is 140 - 145) = 11
$f$ (frequency of the median class 145 - 150) = 18
$h$ (class size) = 5

Substitute these values into the formula:

Median = 145 + (\frac{25.5 - 11}{18}) \times 5

Median = 145 + (\frac{14.5}{18}) \times 5

Median = 145 + \frac{72.5}{18}

Median = 145 + 4.027...

Median \approx 149.03

Answer: The median height is approximately 149.03 cm.

Remember
When dealing with "less than" or "more than" type cumulative frequency distributions, the first step is always to convert them into a standard grouped frequency distribution with defined class intervals and their corresponding frequencies. The median class is determined by finding the class interval whose cumulative frequency is the first to be greater than or equal to $\frac{n}{2}$ .

Exam-smart
In a case study or data interpretation question, you might be given a cumulative frequency table (less than or more than type) and asked to find the median. You will need to reconstruct the standard frequency table, calculate $n$ , find $\frac{n}{2}$ , identify the median class, and then apply the median formula. Alternatively, you could be given the median and some frequencies and asked to find missing frequencies, as shown in Example 2.

Try this: The following table shows the distribution of 50 students according to the marks obtained by them:

Marks obtained	Number of students
0 - 10	5
10 - 20	15
20 - 30	10
30 - 40	8
40 - 50	12
Find the median marks.
Answer: 27.5

Statistics — Notes

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Introduction to Statistics

Measures of Central Tendency for Grouped Data

Mean of Grouped Data

1. Direct Method

Example 1 (Ungrouped to Grouped Data)

Example 2 (Grouped Data)

2. Assumed Mean Method

Example 2 (Assumed Mean Method)

3. Step-Deviation Method

Example 2 (Step-Deviation Method)

Comparison of Methods for Mean

Example 2 (All Three Methods)

Example 3 (Unequal Class Sizes)

13.3 Mode of Grouped Data

Example 4 (Ungrouped Data)

Example 5 (Grouped Data)

Example 6 (Mode and Mean Comparison)

Median of Grouped Data

Finding the Median Class

Formula for Median of Grouped Data

Example 1: Calculating Median from a Grouped Frequency Distribution

Example 2: Finding Unknown Frequencies using Median

Example 3: Median from "Less Than" Type Data

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